#leetcode题目97：交错字符串
#难度：中等

class Solution:
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        m = len(s1)
        n = len(s2)
        t = len(s3)
        if m + n != t:
            return False
        dp = [[False] * (n + 1) for _ in range(m + 1)]
        dp[0][0] = True
        for i in range(1, m + 1):
            dp[i][0] = dp[i-1][0] and s1[i-1] == s3[i-1]
        for j in range(1, n + 1):
            dp[0][j] = dp[0][j-1] and s2[j-1] == s3[j-1]
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                dp[i][j] = (dp[i-1][j] and s1[i-1] == s3[i+j-1]) or (dp[i][j-1] and s2[j-1] == s3[i+j-1])
        return dp[-1][-1]


#测试数据
s1 = "aabcc"
s2 = "dbbca"
s3 = "aadbbcbcac"
#预期输出：True
solution = Solution()
print(solution.isInterleave(s1, s2, s3))

s1 = "aabcc"
s2 = "dbbca"
s3 = "aadbbbaccc"
#预期输出：False
solution = Solution()
print(solution.isInterleave(s1, s2, s3))

s1 = ""
s2 = ""
s3 = ""
#预期输出：True
solution = Solution()
print(solution.isInterleave(s1, s2, s3))

